Two perpendicular lines are given in a coordinate system intersecting at A(6,8). The points P and Q are the intersections of those lines with the y-axis and they are symmetric with respect to the origin. Find the area of the triangle APQ.
Answers
Answer:
Here 0P = OQ as lines are symmetric around Origin.
Check the attached image for the complete solution.
Given:
two perpendicular lines intersecting at A(6,8).
P and Q are the intersections of A with the y- axis
they are symmetric.
To Find:
the area of the triangle APQ.
Solution:
OP = OQ (the lines are symmetric concerning the origin)
In ΔAMP,
⇒ (AM)² + (PM)² = (AP)² [ using Pythagoras theorem]
⇒ (AP)² = x² + 36 ..(i)
In ΔAMQ,
⇒ (AM)² + (MQ)² = (AQ)²
⇒ 36 + [X + 8 + 8]² = (AQ)²
⇒ (AQ)² = x² + 32x +256 + 36
⇒ (AQ)² = x² + 32x + 292 ..(ii)
Now,
⇒ (AP)² + (AQ)² = (PQ)² [ lines are perpendicular]
⇒ (x² + 36) + (x² +32x+ 292) = (x + 8 + x + 8)²
⇒ 2x² + 32x + 328 = (2x+ 16)²
⇒ 2x² +32x+ 328 = 4x² + 64x + 256
⇒ 2x²+ 32x-72 = 0
⇒ x² + 16x-36 = 0
⇒ x = 2
Then,
area of ΔAPQ = 1/2×b×h
= 1/2 × (PQ) × (AM)
= 1/2 × 20 × 6
= 60² units.
Therefore, the area of ΔAPQ is 60²units.